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It seems quite obvious that, when given a simplex, its set of vertices is uniquely determined by the simplex. The formal formulation of this intuition is as follows:

Suppose that the points $\{v_0,v_1,\dots,v_k\}$ and $\{w_0,w_1,\dots,w_l\}$ are affinely independent sets of points of $\mathbb{R}^n$. If the $k$-simplex $\sigma=[v_0,\dots,v_k]$ and the $l$-simplex $\tau=[w_0,\dots, w_l]$ are equal, then $k=l$ and $\{v_0,v_1,\dots,v_k\}=\{w_0,w_1,\dots,w_l\}$.

We clearly have $k=l$ because an $m$-simplex is homeomorphic to the closed ball of dimension $m$ and no two closed ball of different dimensions are homeomorphic to each other. But I cannot prove that $\{v_0,v_1,\dots,v_k\}=\{w_0,w_1,\dots,w_l\}$.

I think I must be overlooking something very obvious... Can anyone help me? Thanks in advance.

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  • $\begingroup$ What does it mean for two simplices to be equal? $\endgroup$ – Connor Malin Mar 27 '19 at 3:46
  • $\begingroup$ @ConnorMalin Two simplices are equal when they are equal as sets. $\endgroup$ – Ken Mar 27 '19 at 4:04
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    $\begingroup$ Given a simplex, you can recover its vertices by its extreme points. Maybe this helps. $\endgroup$ – JHF Mar 27 '19 at 16:33
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Trickier than I expected. Suppose that $\{v_0,\dots,v_k\}$ does not equal $\{w_0,\dots,w_k\}$. Then without loss of generality we can assume $v_0 = \Sigma r_i w_i$ where multiple $r_i$ are nonzero. Again, without loss of generality assume that these are $r_0,r_1$. Then there is a function $(-\epsilon, \epsilon) \rightarrow \sigma$ defined by $t \rightarrow (r_0 +t)w_0 + (r_1-t)w_1 + r_2 w_2 +\dots + r_k w_k$. Such a function cannot exist because at $t=0$ this passes through $v_0$, and no such line segment passes through $v_0$ that is also contained in $\sigma$ because such a thing would necessarily contain points that when written as a sum of the $v_i$ would have negative coefficients.

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As JHF had pointed out in the comment, the vertices of a given simplex can be characterized as the extreme points of the simplex. (If $S$ is a convex set, a point $x$ in $S$ is called its extreme point if for each $y$ in $S\setminus \{x\}$, any open line segment containing $x$ and $y$ always contains a point not in $S$.) This can be checked directly by using the barycentric coordinates.

Although the underlying idea is the same as that of Connor Malin's answer, and despite this answer being not my idea, I will post this because I feel like this is the most "natural" way to approach the problem. Big thanks to Connor Malin and JHF!

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